If kerosene, coloured saline water and water are taken in a bottle ,
How are they arranged? give reason
Answers
Explanation:
10th term of an A.P is \sf \dfrac{1}{20}
20
1
and it's 20th term is \sf \dfrac{1}{10}
10
1
.
\Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: Find:-}}}}}}}†
ToFind:−
Sum of the first 200 terms of the A.P.
\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Solution:-}}}}}}}†
Solution:−
First,
10th term of the A.P = \sf \dfrac{1}{20}
20
1
This implies that,
\sf a_{10} \: = \: \dfrac{1}{20}a
10
=
20
1
a + 9d = \sf \dfrac{1}{20} \longrightarrow \boxed{1}
20
1
⟶
1
Now,
20th term of the A.P = \sf \dfrac{1}{10}
10
1
This implies that,
\sf a_{20} \: = \: \dfrac{1}{10}a
20
=
10
1
a + 19d = \sf \dfrac{1}{10} \longrightarrow \boxed{2}
10
1
⟶
2
Now,
\boxed{2} \: - \: \boxed{1}
2
−
1
,
a + 19d = \sf \dfrac{1}{10}
10
1
- \bigg(( a + 9d = \sf \dfrac{1}{20}
20
1
\bigg))
Then,
\sf \cancel{a}
a
+ 19d = \sf \dfrac{1}{10}
10
1
\sf \cancel{- \: a}
−a
- 9d = \sf - \: \dfrac{1}{20}−
20
1
————————————
10d = \sf \dfrac{1}{10} - \dfrac{1}{20}
10
1
−
20
1
Then,
10d = \sf \dfrac{2}{20} - \dfrac{1}{20}
20
2
−
20
1
10d = \sf \dfrac{2 - 1}{20}
20
2−1
10d = \sf \dfrac{1}{20}
20
1
d = \sf \dfrac{1}{20 * 10}
20∗10
1
d = \sf \dfrac{1}{200}
200
1
Answer:
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Explanation:
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